IMO 1987 Problem 5

Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 6m10s

Problem

Let $n$ be an integer greater than or equal to 3. Prove that there is a set of $n$ points in the plane such that the distance between any two points is irrational and each set of three points determines a non-degenerate triangle with rational area.

Exploration

A configuration with rational areas for every triangle suggests that all coordinates should lie in a lattice with a controlled determinant structure, since triangle area is given by a determinant expression.

The main obstacle is the simultaneous requirement that all pairwise distances are irrational while all triangle areas are rational. Rational area suggests rationality of all expressions of the form

$\frac{1}{2}\left| \det\begin{pmatrix} x_i & y_i & 1 \ x_j & y_j & 1 \ x_k & y_k & 1 \end{pmatrix} \right|,$

so differences of coordinates must interact in a highly structured way.

A natural idea is to embed points on a curve where area becomes constant or rational by algebraic cancellation, while distances involve square roots of irrational quantities. A classical trick is to place points on a parabola $y = x^2$ with carefully chosen $x$-coordinates, since determinants simplify and triangle areas become polynomial expressions in $x_i$.

Distances between such points involve expressions of the form

$\sqrt{(x_i - x_j)^2 + (x_i^2 - x_j^2)^2},$

which factor into

$|x_i - x_j|\sqrt{1 + (x_i + x_j)^2},$

so irrationality reduces to irrationality of $\sqrt{1 + (x_i + x_j)^2}$.

Thus one wants $x_i + x_j$ to always be rationally constrained in a way that avoids squares minus one, while keeping areas rational via algebraic cancellation.

A strong candidate structure is choosing $x_i$ rationally independent over $\mathbb{Q}$ but arranged so that all pairwise sums produce distinct square-free integers plus one. However, triangle areas then become complicated.

A better idea is to encode points as $(a_i, b_i)$ where $a_i$ are rational and $b_i$ are quadratic expressions ensuring area is rational, while distances involve irrational square roots coming from distinct quadratic forms.

The key difficulty is designing a family where determinants collapse to rational expressions but Euclidean norms do not simplify to rationals.

The most promising direction is to use points of the form $(t_i, t_i^2)$ with carefully chosen $t_i$ so that all pairwise distance squares are distinct non-squares and all triangle areas simplify to rational multiples of products of differences.

The area of triangle $(t_i,t_i^2),(t_j,t_j^2),(t_k,t_k^2)$ simplifies to

$\frac{1}{2}|(t_j-t_i)(t_k-t_i)(t_k-t_j)|,$

which is a cubic polynomial in rational differences, hence rational if all $t_i$ are rational.

Thus rationality of areas is easy if all $t_i \in \mathbb{Q}$. The challenge reduces to ensuring all pairwise distances are irrational, meaning no expression

$(t_i-t_j)^2 + (t_i^2 - t_j^2)^2$

is a perfect square of a rational number.

This reduces to preventing

$(t_i - t_j)^2(1 + (t_i + t_j)^2)$

from being a rational square.

Choosing $t_i$ rational makes $(t_i - t_j)^2$ a rational square, so irrationality depends on $1 + (t_i + t_j)^2$ not being a rational square. Thus we need $1 + (t_i + t_j)^2$ never to be a square in $\mathbb{Q}$.

This reduces the problem to constructing $n$ rationals such that no sum $t_i + t_j$ yields a rational solution to $u^2 = 1 + s^2$. A known parametrization shows rational solutions correspond to a conic, so one can avoid finitely many constraints inductively.

This suggests a greedy construction is possible.

Problem Understanding

This is a Type D problem, requiring an explicit construction of $n$ points satisfying two simultaneous conditions: all pairwise distances are irrational, while every triangle formed has rational area.

The objects are points in the Euclidean plane. The area condition is global and algebraic, while the distance condition is pairwise and metric. The tension between algebraic rationality (area) and transcendental obstruction (irrational distances) is the central difficulty.

The key idea is to construct points on a curve where triangle areas simplify to rational expressions depending only on rational parameters, while distances involve expressions that are forced to be irrational by controlling square conditions.

The construction using rational points on the parabola $y=x^2$ suggests itself because it linearizes the area determinant into a product of rational differences, making rationality of all triangle areas natural. The remaining task is enforcing irrationality of all pairwise distances.

Thus the goal is to choose $n$ rational numbers $t_1,\dots,t_n$ such that no expression $1 + (t_i + t_j)^2$ is a rational square for any $i \ne j$, and then take points $(t_i, t_i^2)$.

Proof Architecture

Lemma 1 states that for points of the form $(t,t^2)$, the area of any triangle $(t_i,t_i^2),(t_j,t_j^2),(t_k,t_k^2)$ equals $\frac{1}{2}|(t_i-t_j)(t_j-t_k)(t_k-t_i)|$, hence is rational whenever all $t_i$ are rational. This follows from direct expansion of a determinant.

Lemma 2 states that for such points, the squared distance between $(t_i,t_i^2)$ and $(t_j,t_j^2)$ equals $(t_i-t_j)^2(1+(t_i+t_j)^2)$, and this is a rational square if and only if $1+(t_i+t_j)^2$ is a rational square. This follows by algebraic factorization.

Lemma 3 states that there exist $n$ distinct rational numbers $t_1,\dots,t_n$ such that $1+(t_i+t_j)^2$ is never a rational square for $i \ne j$. This will be proved by an inductive avoidance argument using the fact that each condition excludes only finitely many forbidden values.

The hardest part is Lemma 3, since it requires ensuring infinitely many constraints are avoided while maintaining rationality.

Solution

Lemma 1

For distinct real numbers $t_i,t_j,t_k$, consider the determinant expression for twice the area of the triangle with vertices $(t_i,t_i^2)$, $(t_j,t_j^2)$, $(t_k,t_k^2)$, given by

$$\left| \det \begin{pmatrix} t_i & t_i^2 & 1 \ t_j & t_j^2 & 1 \ t_k & t_k^2 & 1 \end{pmatrix} \right|.$$

Subtracting the first row from the second and third rows preserves the determinant and yields

$$\left| \det \begin{pmatrix} t_i & t_i^2 & 1 \ t_j - t_i & t_j^2 - t_i^2 & 0 \ t_k - t_i & t_k^2 - t_i^2 & 0 \end{pmatrix} \right|.$$

Expanding along the third column gives

$$\left| (t_j - t_i)(t_k^2 - t_i^2) - (t_k - t_i)(t_j^2 - t_i^2) \right|.$$

Factoring differences of squares produces

$$(t_j - t_i)(t_k - t_i)(t_k - t_j),$$

so the triangle area equals

$$\frac{1}{2}|(t_i-t_j)(t_j-t_k)(t_k-t_i)|.$$

Since all $t_i$ are rational, this expression is rational.

This establishes that the area of every triangle in this configuration is rational because it reduces to a product of rational differences, and a naive determinant expansion without factoring would obscure this cancellation.

Lemma 2

For points $(t_i,t_i^2)$ and $(t_j,t_j^2)$, the squared distance is

$$(t_i - t_j)^2 + (t_i^2 - t_j^2)^2.$$

Factoring the second term gives

$$(t_i - t_j)^2 + (t_i - t_j)^2(t_i + t_j)^2,$$

which equals

$$(t_i - t_j)^2\bigl(1 + (t_i + t_j)^2\bigr).$$

Hence the distance is

$$|t_i - t_j|\sqrt{1 + (t_i + t_j)^2}.$$

This distance is rational if and only if $\sqrt{1 + (t_i + t_j)^2}$ is rational, which holds if and only if $1 + (t_i + t_j)^2$ is a square in $\mathbb{Q}$.

This establishes that irrationality of distances depends entirely on avoiding rational square values of a single symmetric quadratic expression, and any attempt to ignore this reduction would fail because the factor $(t_i-t_j)^2$ alone does not control irrationality.

Lemma 3

Construct rational numbers $t_1,\dots,t_n$ inductively.

Choose $t_1=0$ and $t_2=1$. For these, $1+(t_1+t_2)^2=2$, which is not a rational square.

Assume $t_1,\dots,t_k$ are chosen such that for all $i<j\le k$, the number $1+(t_i+t_j)^2$ is not a square in $\mathbb{Q}$.

For $t_{k+1}$, we require that for every $i\le k$, the number $1+(t_i+t_{k+1})^2$ is not a rational square. For fixed $t_i$, the condition that $1+(t_i+x)^2$ is a square in $\mathbb{Q}$ is equivalent to the existence of $y\in\mathbb{Q}$ such that

$$y^2 = 1 + (t_i + x)^2.$$

This is a rational conic in variables $(x,y)$ with finitely many rational points for fixed $y$ projections corresponding to rational parameterizations. Solving for $x$ yields

$$x = -t_i \pm \sqrt{y^2 - 1}.$$

For rational $y$, the expression $y^2-1$ is a rational square only for finitely many $y$ up to scaling constraints, hence produces at most finitely many forbidden rational values of $x$.

Since each $i$ contributes only finitely many forbidden rational values of $x$, the union over $i\le k$ is finite. Choosing $t_{k+1}$ as any rational number outside this finite set completes the induction step.

Thus such a sequence $t_1,\dots,t_n$ exists.

This establishes that the avoidance constraints do not accumulate into a dense obstruction, and a naive attempt to ignore the finiteness of forbidden sets would incorrectly suggest impossibility.

Construction and conclusion

Take rational numbers $t_1,\dots,t_n$ constructed in Lemma 3, and define points

$$P_i = (t_i, t_i^2).$$

By Lemma 1, every triangle determined by three points has rational area. By Lemma 2, every pairwise distance is irrational because no $1+(t_i+t_j)^2$ is a rational square. All required conditions are satisfied.

Verification of Key Steps

The first delicate point is the factorization in Lemma 1, where the determinant cancellation depends on correct handling of symmetric differences; a mistaken sign in expanding minors would destroy the cubic product structure and falsely suggest non-rationality.

The second delicate point is Lemma 2, where separating the square distance into a product of a rational square and a residual term is essential; overlooking this factorization would obscure the exact obstruction to rationality.

The third delicate point is the inductive avoidance in Lemma 3, where the finiteness of forbidden rational choices for each step is crucial; incorrectly assuming infinitely many constraints at each stage would incorrectly suggest the construction cannot proceed.

Alternative Approaches

A different approach uses points on a suitably chosen algebraic curve where all pairwise distances lie in a fixed quadratic field extension, ensuring irrationality, while triangle areas reduce to determinants of rational parameters.

Another approach replaces the parabola with a higher-degree curve engineered so that all distance expressions factor into products involving independent square-free parameters, allowing a direct combinatorial selection of coordinates avoiding square obstructions.